# Ph pka and pkb relationship test

### pH and pKa relationship for buffers (video) | Khan Academy

Henderson equation best relates these two terms.. take it easy don't overthink. pKa = -log(Ka) and so we get an equation relating pH and pKa: pH = -log(Ka) + log([HA]/[A-]). So, the only way to After the test they forget. So did I, but since I. equation. This equation is very useful in calculating the pH of a solution containing a pOH = pKb + log ([Salt]/[Base]). constant + log ([Salt]/[Acid]), where the constant is the pKa of the particular weak acid used in the . these measurements of pH, the combination electrode will be inserted into each of the test tubes, and. Regarding the last equation, it is only applicable to the sum of the pKa of a species HX and the pKb of its conjugate base, X-. Since both the pK.

So it's going to be plus-- oh, this is in an aqueous solution. Let me do that. And then you're going to have the rest of the acid, whatever it might be. A minus, and that's also going to be in an aqueous solution. And that's the general pattern. We've seen the case where A could be an NH3, right? And this is just NH3. A could be a fluorine molecule right there, because then this would be hydrogen fluoride or hydrofluoric acid.

And this would just be the negative ion of fluorine. Or a fluorine with an extra electron. So it could be a bunch of stuff. You can just throw in anything there, and it'll work. Especially for the weak acids. So we learned a last video that if this is the acid, then this is the conjugate base. And we could write the same reactions, essentially, as kind of more of a basic reaction. So we could say, if I start with A minus-- it's in an aqueous solution-- that's in equilibrium with-- This thing could grab a hydrogen from the surrounding water and become neutral then.

It's still in an aqueous solution. And then one of those water molecules that it plucked that hydrogen off of is now going to be a hydroxide molecule. Remember, whenever I say pluck the hydrogen, just the proton, not the electron for the hydrogen. So the electron stays on that water molecule, so it has a negative charge.

### Oxford University Press | Online Resource Centre | Multiple choice questions

It's in an aqueous solution. So we could write the same reaction both ways.

And we can write equilibrium constants for both of these reactions. So let's do that. Let me erase this, just because I can erase this stuff right there, and then use that space. So an equilibrium reaction for this first one. I could call this the K sub a, because the equilibrium reaction for an acid. And so this is going to be equal to its products. So the concentration of my hydrogen times my concentration of whatever my conjugate base was, divided by my concentration of my original acid.

So this would be the concentration of HA. I could also write an equilibrium constant for this basic reaction.

Let me do it right down here. So I'll call that my K sub b. This is a base equilibrium. And so this is equal to the concentration of the products. It's becoming tedious to keep switching colors. Actually, I'll do it. Because it makes it easier look at, at least for me. HA times the concentration of my hydroxide ions divided by my concentration of my weak base. Remember, this can only be true of a weak base or a weak acid. If we were dealing with a strong acid or is or a strong base, this would not be an equilibrium reaction.

It would only go in one direction. And when it only goes in one direction, writing this type of equilibrium reaction makes no sense-- or equilibrium constant-- because it's not in equilibrium. It only goes in one direction. If A was chlorine, if this was hydrochloric acid, you couldn't do this.

## Chapter 17: Multiple choice questions

You would just say look, if you have a mole of this, you're just dumping a mole of hydrogen protons in that solution and then a bunch of chlorine anions who are not going to do anything. Even though they are the conjugate base, they wouldn't do anything. So you can only do this, remember, for weak acids and bases.

So with that said, let's see if we can find a relationship between Ka and Kb. What do we have here? We have an A minus on both sides of this. We have H over-- OH over A minus. Let's solve for A minus. If we multiply both sides of this equation by HA over H plus, on the left-hand side we get Ka times the inverse of this. So you have your HA over H plus is equal to your concentration of your conjugate base.

And let's do the same thing here. Solve for A minus. As you can see from these plots, the titration curve for adding a base is the mirror image of the curve for adding an acid. Suppose that we now add 0. Thus the pH of the solution increases gradually.

For the titration of a monoprotic strong acid HCl with a monobasic strong base NaOHwe can calculate the volume of base needed to reach the equivalence point from the following relationship: At the equivalence point when The pH is initially As the equivalence point is approached, the pH drops rapidly before leveling off at a value of about 0.

- 17.3: Acid-Base Titrations
- pH and pKa relationship for buffers
- pH, pKa, Ka, pKb, Kb

The titration of either a strong acid with a strong base or a strong base with a strong acid produces an S-shaped curve. The curve is somewhat asymmetrical because the steady increase in the volume of the solution during the titration causes the solution to become more dilute. Due to the leveling effect, the shape of the curve for a titration involving a strong acid and a strong base depends on only the concentrations of the acid and base, not their identities.

The shape of the titration curve involving a strong acid and a strong base depends only on their concentrations, not their identities. So anything to the zeroth power is equal to one.

Which tells us that this ratio is equal to one. And if A minus concentration over HA concentration is equal to one, that means that they have the same concentration.

### The pKa Table Is Your Friend — Master Organic Chemistry

I forgot a minus sign there. This is a really helpful thing to remember. And this comes up a lot not just when you're talking about buffers by themselves, but also when you're doing titrations. And the point in your titration where the HA is equal to A minus is called the half-equivalence point.

And if you haven't learnt about buffers, that's okay. Oh, sorry, if you haven't learn about titrations yet, that's fully fine. Just ignore what I just said laughsbut if you have, the moral is just that, this is a really, really important relationship that is really helpful to remember.

And I said really a lot there. And when you raise 10 to a positive number, when you raise 10 to a positive number, you get a ratio that is greater than one.